Only smart people are allowed to answer ONLY smart people............... If you can answer this correctly then maybe you are actually yourself!!!Fill in the blanks: ______ compounds break up into ______ and ______ in solution, which can move freely and carry an _______ current.

Answer: The mass of molten iron formed will be 1.92 kg

Explanation:

We are given:

Moles of iron (III) oxide = 17.2 moles

To calculate the number of moles, we use the equation:

    ......(1)

Given mass of aluminium = 2.28 kg = 2280 g     (Conversion factor:  1 kg = 1000 g)

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of aluminium and iron (III) oxide follows:

By Stoichiometry of the reaction:

1 mole of iron (III) oxide reacts with 2 moles of aluminium

So, 17.2 moles of iron (III) oxide will react with = of aluminium

As, given amount of aluminium is more than the required amount. So, it is considered as an excess reagent.

Thus, iron (III) oxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of iron (III) oxide produces 2 moles of iron

So, 17.2 moles of iron (III) oxide will produce =   of iron

Now, calculating the mass of iron by using equation 1, we get:

Molar mass of iron = 55.8 g/mol

Moles of iron = 34.4 moles

Putting values in equation 1, we get:

Hence, the mass of molten iron formed will be 1.92 kg

Answer:

1. D. cal......

2.A. iron

3. D

4.2.44j/g°C   A

5,Lf=334J/g   B

Explanation:

1: Which of the following is the abbreviation for a unit of energy? A. K / B. °C/ C. W / D. cal...............

calorie is the unit of energy

#2: A 200 g block of a substance requires 1.84 kJ of heat to raise its temperature from 25°C to 45°C. Use the table attached to identify the substance. A. iron/ B. aluminum/ C. gold/ D. copper.....................

Q=mcdt

1840=0.2*C*(45-25)

C=460J/KgK

if the specific heat capacity is the above then he substance is iron

#3: In a calorimeter, the temperature of 100 g of water decreased by 10°C when 10 g of ice melted. How much heat was absorbed by the ice? A. 418 kJ / B. 100 kJ / C. 10 J / D. 4.18 kJ .................

Q=mcdT

Q=0.1*10*4180

Q=4180j. answer D

.#4: The amount of heat needed to raise the temperature of 50 g of a substance by 15°C is 1.83 kJ. What is the specific heat of the substance? A. 2.44 J/g-°C / B. 2.22 J/g-°C / C. 2.13 J/g-°C / D. 2.05 J/g-°C ................

Q=mcdT

1830=50/1000*C*15

C=2440j/kg/k

change it to j/g°C

2.44j/g°C   A

#5: In a calorimeter, 3.34 kJ of heat was absorbed when 10 g of ice melted. What is the enthalpy of fusion of the ice? A. 6.68 J/g / B. 334 J/g / C. 6.68 kJ/g/ D. 334 kJ/g

Q=mLf

Lf=enthalpy of fusion

3340/10=Lf

Lf=334J/g   B

Enthalpy of fusion quantity of heat to convert 1 unit mass of a solid to liquid without any noticeable change in temperature.

answer: THIS IS CORRECT I TOOK THE TEST D: BOTH A AND C

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Answer: The value of enthalpy of the given reaction is 44.000 kJ/mol , that is energy is supplied to water to change into water vapors

Explanation:

Enthalpy of formation of water in liquid state,

Enthalpy of formation of water in gaseous state,

The value of enthalpy of the given reaction is 44.000 kJ/mol, that is energy is supplied to water to change into water vapors.